We were discussing Master Data management and how it differs between MongoDB and traditional databases. MongoDB provides two important functionalities over catalog namely browsing and searching. Both these services seem that they could do with improvements. While browsing based service could expand on the standard query operator capabilities including direct SQL invocations, the search capabilities could be made similar to that of Splunk which can query events using search expressions and operators that have the same look and feel as customary tools on Unix. These two distinct services are provided over a single managed view of the catalog. The challenge with browsing over catalog unlike other datasets is that the data is quite large and difficult to navigate. Traditional REST APIs solve this with the use of a few query parameters such as page, offset, search-term and leaving a cursor like logic to the callers. Instead I propose to push the standard query operators to service itself.
The browsing service has to be optimized for the mobile experience as compared to the desktop experience. These optimizations include smaller page size, reorganized page layouts, smaller resources, smaller images and  bigger thumbnails, mobile optimized styling, faster page load times and judicious use of properties on the page.
Since much of the content for catalog remain static resources, web proxies, gro sharding, content delivery networks, web proxies and app fabric like caches are immensely popular. However, none of these are really required if the data store is indeed cloud based.  For example the service level agreement from a cloud database is the same regardless of the region from where the query is issued.
#codingexercise
we were discussing how to find whether a given tree is a sub tree of another tree. We gave a recursive solution and an iterative solution.  The recursive solution did not address the case when the subtree is not a leaf in the original tree. Since we know that the original and subtree are distinct we can improve  the recursive solution by adding a condition that returns true if the subtree node is null and the original is not.

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bool isEqual(Node root1, Node root2)
{
if (root1 == NULL && root2 == NULL) return true;
if (root1 == NULL) return false;
if (root2 == NULL) return true; // modified
return root1.data == root2.data && isEqual(root1.left, root2.left) && isEqual(root1.right, root2.right);
}